The joint probability distribution of thr random variables X and Y is given below: f(x,y)={(cxy , 0<x<2,0<y<x),(0,, \text{ other }):}

The joint probability distribution of thr random variables X and Y is given below:

a.Find the value of the constant.
b.Calculate the covariance and the correlation of the X and Y random variables.
c. Calculate the expected value of the random variable $Z=2X-3Y+2$

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Step 1
Given:

a. value of constant c,
${\int }_{0}^{2}{\int }_{0}^{x}cxydydx=1$
${\int }_{0}^{2}cx{\left[\frac{{y}^{2}}{2}\right]}_{0}^{x}dx=1$
$\frac{c}{2}{\int }_{0}^{2}x×{x}^{2}dx=1$
$\frac{c}{2}{\left[\frac{{x}^{4}}{4}\right]}_{0}^{2}=1$
2c=1
$c=\frac{1}{2}$
Step 2
b.
we have given a joint pdf,

marginal pdf of x,
${f}_{1}\left(x\right)={\int }_{0}^{x}f\left(x,y\right)dy$
${f}_{1}\left(x\right)={\int }_{0}^{x}\frac{1}{2}xydy$
${f}_{1}\left(x\right)=\frac{1}{2}x×{\left(\frac{{y}^{2}}{2}\right)}_{0}^{x}$

$E\left(x\right)={\int }_{0}^{2}x×{f}_{1}\left(x\right)dx$
$E\left(x\right)={\int }_{0}^{2}×\frac{{x}^{3}}{4}dx$
$E\left(x\right)=\frac{1}{4}{\left(\frac{{x}^{5}}{5}\right)}_{0}^{2}$
$E\left(x\right)=\frac{8}{5}$
Step 3
$E\left({x}^{2}\right)={\int }_{0}^{2}{x}^{2}×{f}_{1}\left(x\right)dx$